Recitation 2 (Chapter 2) Jan. 24-28 Ch. 2 # 9. A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.0 m/s. The car is a distance d away. The bear is 26 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d ? REASONING In order for the bear to catch the tourist over the distance d , the bear must reach the car at the same time as the tourist. During the time t that it takes for the tourist to reach the car, the bear must travel a total distance of d + 26 m. From Equation 2.1, tourist d v t = (1) and bear 26 m d v t + = (2) Equations (1) and (2) can be solved simultaneously to find d . SOLUTION Solving Equation (1) for t and substituting into Equation (2), we find tourist bear tourist ( 26 m) 26 m / d v d v d v d + + = = bear tourist 26 m 1 v v d   = +     Solving for d yields: bear tourist 26 m 26 m 52 m 6.0 m/s 1 1 4.0 m/s d v v = = =--______________________________________________________________________________ Ch. 2 # 18. A sprinter explodes out of the starting block with an acceleration of +2.3 m/s 2 , which she sustains for 1.2 s. Then, her acceleration drops to zero for the rest of the race. (a) What is her velocity at t = 1.2 s. (b) What is her velocity at the end of the race? REASONING We can use the definition of average acceleration ( 29 ( 290 / t t =--0a v v (Equation 2.4) to find the sprinter’s final velocity v at the end of the acceleration phase, because her initial velocity ( 0 m/s =0 v , since she starts from rest), her average acceleration a , and the time interval t-t0 are known. SOLUTION a. Since the sprinter has a constant acceleration, it is also equal to her average acceleration, so 2 2.3 m/s = + a Her velocity at the end of the 1.2-s period is ( 29 ( 29 ( 29 ( 29 20 0 m/s 2.3 m/s 1.2 s 2.8 m/s t t = +-

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